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Volumen 24, 1999, 465–488

UPPER SETS AND QUASISYMMETRIC MAPS

D.A. Trotsenko1 and J. V¨ais¨al¨a

Academy of Sciences, Institute of Mathematics

Universitetskij prospekt 4, 630090 Novosibirsk, Russia; trotsenk@math.nsc.ru University of Helsinki, Department of Mathematics

PL 4, Yliopistonkatu 5, FIN-00014 Helsinki, Finland; jvaisala@cc.helsinki.fi

Abstract. The upper set ˜A of a metric space A is a subset of A×(0,) , consisting of all pairs (x,|xy|) with x, y A, x = y. We consider various properties of ˜A and a metric of ˜A, called the broken hyperbolic metric. The theory is applied to study basic properties of quasisymmetric maps.

1. Introduction

1.1. Let A be a metric space, where the distance between points a, b is written as |a−b|. The upper set of A is the subset

A˜=

(x,|x−y|) :x ∈A, y ∈A\ {x}

of R+, R+ = (0,) . We assume that A contains at least two points in order that ˜A be nonempty. If A⊂Rn, then ˜A Rn+1+ =Rn×R+, and we can consider the hyperbolic metric h of Rn+1+ in ˜A. However, we find it convenient to replace h by another metric , which is bilipschitz equivalent to h, and which is easy to consider also in the case where A is an arbitrary metric space. The precise definition of is given in 2.2.

For each λ > 0 we partition ˜A into λ-components, where points z and z belong to the same λ-component if they can be joined by a finite sequence z = z0, . . . , zN = z in ˜A with (zj1, zj) λ for all j. The family Γ = Γλ(A) of all λ-components of ˜A is the main object of study in this paper. The case λ 1 is the most interesting. There is a naturalordering of Γ , which gives Γ a structure of a treelike graph.

If A is connected, then also ˜A is connected, and Γ is the trivial graph con- sisting of one vertex. More generally, Γ is trivial if A is c-uniformly perfect with c < eλ; see 4.12 for the definition of uniform perfectness. In the general case we can roughly say that the structure of Γ describes in which way A is disconnected.

1991 Mathematics Subject Classification: Primary 30C65.

1 The research described in this paper was supported by the Russian Fund for Fundamental Research (project 96-01-01558), by the University of Helsinki, and by the Finnish Academy of Sciences.

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In Section 6 we apply the theory of upper sets to quasisymmetric maps. Let A and B be metric spaces and let f: A B be η-quasisymmetric; see 6.2 for the definition of quasisymmetry. It is well known that if A is uniformly perfect, then η can be chosen to be of the form η(t) = C(tα∨t1/α) ; see [TV], [AT]. We show that this is true for all η-quasisymmetric maps of A if and only if there is λ such that Γλ(A) is trivial. This property can also be expressed in terms of relative connectedness (Theorem 4.9).

The second author wants to make it clear that an early version of the theory of upper sets was created and developed by the first author D.A. Trotsenko in the late eighties and considered in the conference proceedings [Tr]. Also the application to quasisymmetric maps was proved by him. The main part of this paper was written during the visits of the first author at the University of Helsinki in 1997.

The second author has “polished up” the theory by writing and simplifying details of proofs, introducing some terminology and auxiliary concepts, etc.

1.2. Notation. Throughout the paper, A will denote a metric space with distance between points x, y written as |x−y|. For nonempty sets E, F A, we let d(E) and d(E, F) denote the diameter of E and the distance between E and F. For r > 0 , B(E, r) is the r-neighborhood {x : d(x, E) < r} of E. Moreover, B(x, r) is the open ball with radius r centered at x; the closed ball is B(x, r) . We let R and N denote the set of real numbers and the set of positive integers, respectively, and we write N0 = N∪ {0}. For real numbers s, t we use the notation

s∧t= min(s, t), s∨t = max(s, t).

We make the convention that λ will always denote a real number with λ≥1 . Acknowledgement. We thank Pekka Alestalo for careful reading of the manu- script and for several valuable comments.

2. Broken hyperbolic metric

2.1. Summary. In this section we introduce the broken hyperbolic metric of the space R+ and prove some properties of it.

2.2. Definitions. The ordinary hyperbolic metric h of the half space Rn+1+ = Rn×R+ is defined by the element of length

dh(x, r) = (|dx|2 +dr2)1/2

r ,

where x∈Rn, r R+, n≥1 . We find it convenient to replace h by the metric defined by the element of length

d(x, r) = |dx|+|dr|

r .

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Thus for z, z Rn+1+ we have

(z, z) = inf

γ

γ

|dx|+|dr| r

over all rectifiable arcs γ joining z and z in Rn+1+ . Since dh d 2dh, we have

h(z, z) (z, z)≤√

2h(z, z)

for all z, z Rn+1+ . The metric is the broken hyperbolic metric of Rn+1+ . More generally, let A be an arbitrary metric space, and let z = (x, r), z = (x, r) be points in R+. Choose points y, y Rn with |y−y| = |x−x|. Then the number

(z, z) =

(y, r),(y, r)

is independent of the choice of y and y, and it gives the broken hyperbolic metric of R+. Since each triple of A can be isometrically embedded into R2, we see that indeed is a metric in R+.

Alternatively, the broken hyperbolic metric of R+ can be defined as follows. Let

π: R+ →A, π2:R+ R+

be the projections. By astep we mean a pair of points in R+. A step (z, z) is vertical if π(z) =π(z) , and (z, z) ishorizontal if π2(z) =π2(z) . The hyperbolic length of a vertical step (z, z) is

lh(z, z) =log π2(z) π2(z)

, and the hyperbolic length of a horizontal step (z, z) is

lh(z, z) = |π(z)−π(z)| π2(z) .

A step path in R+ is a finite sequence ¯z = (z0, . . . , zN) of points in R+

such that the step (zj1, zj) is either horizontal or vertical for all 1 j N. The hyperbolic length of ¯z is the number

lhz) = N

j=1

lh(zj1, zj).

In the special case A = Rn, lhz) is the ordinary hyperbolic length of the path consisting of the segmental paths [zj1, zj] , 1 j N. The broken hyperbolic distance between points z, z R+ is defined as

(z, z) = inf

¯ z lhz)

over all step paths ¯z from z to z. It is not difficult to show that the two definitions for are equivalent. However, in the sequel we shall use the second definition; the first one was given to illustrate the connection between and h in Rn+1+ .

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2.3. Geodesics. Let z = (x, r) and z = (x, r) be points in R+. We want to find a geodesic from z to z, that is, a step path ¯z = (z0, . . . , zN) such that z0 =z, zN =z, and lhz) =(z, z) .

Assume that r r and that ¯z is a step path from z to z. Let t = max2(zj) : 0 j N} be the maximal height of ¯z. Then the sum of the hyperbolic lengths of all vertical steps of ¯z is at least log(t/r) + log(t/r) . The corresponding sum for the horizontal steps is at least |x−x|/t. Hence lhz) lhy) , where ¯y is the step path (z, y1, y2, z) with y1 = (x, t) , y2 = (x, t) . We have

lhy) = |x−x|

t + 2 logt−log(rr).

Using elementary calculus we see that the right-hand side attains its minimum at t =|x−x|/2 . We obtain the following result:

2.4. Theorem. Let z = (x, r) and z = (x, r) be points in R+ with r ≤r and |x−x|=s. If r ≤s/2, then the geodesic in the metric from z to z is the step path

z,(x, s/2),(x, s/2), z , and (z, z) = 2 + log s2

4rr. If r ≥s/2, then the geodesic is the step path

z,(x, r), z , and (z, z) = s

r + logr r.

In particular, if r=r ≥s/2 or if x=x, then the geodesic reduces to the single step (z, z).

2.5. Remarks. 1. Observe that the first case of Theorem 2.4 can occur only if (z, z)>2 .

2. In the case A = Rn, the geodesic from z to z can be considered as an ordinary arc γ Rn+1+ . This arc consists of one, two or three line segments, and it lies on the boundary of the unique square Q such that Q and a pair of its sides are perpendicular to Rn, and the center of Q is in Rn. See Figure 1.

Geodesic of the metric in R+n+1

Rn

z

L(x0)

x0

The point z´ of L(x0) is closest to z

Figure 1.

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2.6. Notation. For x ∈A we let L(x) denote the ray {x} ×R+ =π1{x} ⊂ R+.

2.7. Lemma. Let z = (x, r) ∈A×R+ and x0 ∈A. If |x−x0| ≥r, then

z, L(x0)

=

z,(x0,|x−x0|)

= 1 + log(|x−x0|/r).

If |x−x0| ≤r, then

z, L(x0)

=

z,(x0, r)

= |x−x0|/r.

Proof. The lemma follows from 2.4 by elementary calculus. Alternatively, we can argue as follows. Without loss we may assume that A =R and that x0 = 0 . Let z ∈L(x0) be the point closest to z. Thenz must lie on the geodesic between z and (−x, r) , and the lemma follows from 2.4.

2.8. Remark. I f A Rn, then Lemma 2.7 means that the point z of L(x0) closest to z is such that z lies on the boundary of a square with adjacent vertices (x0,0) and z. See Figure 1.

2.9. Lemma. If z = (x, r), z = (x, r)∈A×R+, then |x−x| ≤re(z,z)−1. Proof. Set s=|x−x|. Since 1 + log(s/r)≤s/r, Lemma 2.7 implies that

1 + logs r

z, L(x)

≤(z, z), and the lemma follows.

2.10. Remark. In a recent manuscript [BS], M. Bonk and O. Schramm con- sider a metric in R, defined by the explicit formula

(x, s),(y, t)

= 2 log|x−y|+s∨t

√st .

It is not difficult to show that this metric is bilipschitz equivalent to our metric . I n fact, /2≤≤2

2.

3. Upper sets and λ-components

3.1. Summary. In this section we develop the basic theory of the upper set ˜A of a metric space A. In particular, we study the properties of the family Γλ(A) of the λ-components of ˜A, λ≥1 .

3.2. Definitions. We recall from the introduction that the upper set of a metric space A is

A˜=

(x,|x−y|) :x∈A, y∈A\ {x}

⊂A×R+.

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Recall also that we always assume that A contains at least 2 points, and hence A˜=∅. The broken hyperbolic metric of R+ defines a metric in ˜A.

Recall from 1.2 that we always assume that λ 1 . By a λ-sequence we mean a finite sequence ¯z = (z0, . . . , zN) in ˜A such that (zj−1, zj) λ for all 1≤j ≤N. Two points in ˜A belong to the same λ-component of ˜A if they can be joined by a λ-sequence in ˜A. If there is only one λ-component, ˜A is λ-connected.

The family Γλ(A) of all λ-components of ˜A is a partition of ˜A. For brevity, we shall write Γλ(A) as Γλ or simply as Γ if there is no danger of misunderstanding.

Recall the notation π: R+ →A and π2: R+ R+ for the projections.

We define a partial ordering in R+ by setting z z if π(z) = π(z) and π2(z) ≤π2(z) .

We also define a relation in Γ by setting γ ≤γ if π(γ)⊂π(γ) .

In Theorem 3.4 we collect together several basic properties of Γ . In particular, we show that the relation is a partial ordering of Γ and give two alternative characterizations for this relation. Geometrically, γ ≤γ means that γ lies below γ in R+.

3.3. Notation. I f x, y ∈A and x =y, we set x, y= (x,|x−y|) . Then x, y and y, x are in ˜A, and (x, y,y, x) = 1 by 2.4. Hence the points x, y and y, x lie in the same λ-component. This simple observation is needed in several arguments, and it is the reason for our convention λ≥1 .

3.4. Theorem. Let γ, γ Γ = Γλ(A).

(1) If γ = γ, z γ, z = x, y ∈ γ, and z < z, then y /∈ πγ. Moreover, for each w∈γ we have (z,πw, y) <1 λ and w <πw, y ∈γ.

(2) γ ≤γ if and only if z ≤z for some z ∈γ, z ∈γ.

(3) γ ≤γ if and only if for each z ∈γ there is z ∈γ with z ≤z.

(4) If γ =γ, then either πγ and πγ are disjoint or one of them is a proper subset of the other.

(5) is a partial ordering of Γ.

(6) If z1 ≤z2 ≤z3 are points of A˜ with z1, z3 ∈γ, then z2 ∈γ. (7) If x, y ∈γ, then x, y∈πγ.

(8) πγ =∪{β Γ :β ≤γ}.

(9) If γ contains (x, r) and (x, r), and if r∧r ≤ |x−x|, then γ contains x, x and x, x.

(10) If x0, y ∈πγ and x∈A with |x−x0| ≤eλ|y−x0|, then x∈πγ. (11) If d(πγ) =∞, then πγ =A, and γ is the greatest element of Γ.

(12) If d(πγ)<∞, then B

πγ,(eλ1)d(πγ)

=πγ. (13) If γ < γ, then d(πγ)(eλ1)d(πγ).

(14) πγ is closed and open in A.

(15) If A is separable, then Γ is countable.

Proof. (1) Write z = (x, r) , z = (x, r) , where r = |x−y|. Suppose that w = (u, t) γ. We must show that u = y and that w < u, y ∈ γ. We may assume that (z, w) ≤λ, since the general case follows from this by induction.

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By 2.9 we have |x−u| ≤reλ1, and hence

r−reλ1 ≤ |u−y| ≤r+reλ1.

Since λ < (z, z) = log(r/r) , we have r > reλ, and hence |u−y| ≥ r(eλ eλ1)>0 . Thus the point w =u, y is in ˜A. Moreover,

(z, w) |u−x|

r + log r

r−reλ−1 ≤e1 + log e

e−1 <1≤λ, which yields w ∈γ.

It remains to show that w < w, that is, t <|u−y|. Assume that t≥ |u−y|. Since (w, w) > λ, we then have t > eλ|u −y| > eλ(eλ eλ1)r. On the other hand, (z, w) λ implies that t reλ, and we get the contradiction 1> eλ−eλ1 ≥e−1 . Part (1) is proved.

Observe that in the situation of (1) we have y, x ∈ γ by 3.3. Hence πγ πγ, and parts (2), (3), (4) follow easily. Furthermore, (4) implies (5), and (6) follows from (2) and (5). Part (7) is clear in view of 3.3.

(8) If β, γ Γ and z = x, y ∈ β γ, then x, y πβ πγ by (7), and hence x πγ. Conversely, assume that x, y πγ, x = y. Choose w γ with x = π(w) . Let β Γ be the λ-component of ˜A containing z = x, y. Since x∈πβ∩πγ, we have either β ≤γ or β > γ by (4). The latter case is impossible, since it implies that w < z by (2) and then y /∈πγ by (1). Hence β ≤γ.

(9) We may assume that r ≤ |x−x|. Choose β Γ containing x, x and x, x. Then β ≤γ by (8). Since z ≤ x, x, we have γ ≤β by (2).

(10) By (8) there is β Γ with x0, y ∈β ≤γ. By (3) there isz = (x0, r)∈γ with x0, y ≤z. I f x0, x ≥ z, then

(x0, x, z) = log |x−x0| r ≤λ,

and hence x0, x ∈ γ, which gives x πγ by (7). If x0, x ≤ z, then x0, x ∈ β ≤γ for some β Γ by (2), and hence x∈πβ ⊂πγ.

(11) follows directly from (10).

(12) Observe that d(πγ)>0 by (7). Suppose that x∈B

πγ,(eλ1)d(πγ) . We can choose ε >0 and points x1, x0, y∈πγ such that

|x−x1| ≤(eλ1)d(πγ)−ε,

|x0−y| ≥d(πγ)−εeλ. Then

|x−x0| ≤ |x−x1|+|x1−x0| ≤(eλ1)d(πγ)−ε+d(πγ)≤eλ|y−x0|, which yields x∈πγ by (10).

(13) and (14) follow from (12).

(15) If A is separable, then so is R+. Since the neighborhoods B(γ, λ/2) of γ Γ are disjoint, Γ is countable.

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3.5. Remark. I f A Rn is infinite, then ( ˜A, ) is unbounded. Indeed, A contains a sequence (xj) of distinct points converging to a point in Rn or to . Then (z1, zj)→ ∞, where zj =xj, xj+1 in the first case, and zj = x1, xj+1 in the second case.

This result is not true for arbitrary metric spaces. For example, if A is a space with the discrete metric |x −y| = 1 for x = y, then ˜A = A× {1}, and (z, z) = 1 for all z =z.

4. Simple λ-sequences and relative connectedness

4.1. Summary. We show that each λ-sequence (z0, . . . , zN) in the upper set ˜A can be embedded into a λ-sequence of a special kind, called asimple λ-sequence.

It turns out that the simple λ-sequences of ˜A are in one-to-one correspondence with certain sequences (x0, . . . , xN) of A, called M-relative sequences and char- acterized by the property

|xj1 −xj|/M ≤ |xj−xj+1| ≤M|xj1 −xj|,

M = eλ. These lead to relatively connected metric spaces, which are closely connected with uniformly perfect spaces.

4.2. Simple sequences. We say that a pair (z, z) of points in the upper set ˜A of a metric space A is simple if π2(z) =π2(z) = |π(z)−π(z)|. I n other words, there are x, y ∈A such that z =x, y, z =y, x. Recall that (z, z) is vertical if π(z) =π(z) . In particular, a pair (z, z) is always vertical (but never simple).

Observe that (z, z) = 1 for a simple pair (z, z) and that (z, z) =log π2(z)

π2(z) for a vertical pair.

A finite sequence ¯z = (z0, . . . , zN) in ˜A is called simple if N is odd and if the step (zj1, zj) is simple for all odd j and (zj1, zj) is vertical for all even j. We first show that every λ-sequence in ˜A can be simplified to a simple λ- sequence by adding new points.

4.3. Theorem. Every λ-sequence z¯= (z0, . . . , zN) of A˜ can be embedded into a simple λ-sequence u¯= (u0, . . . , uN). This means that u0= z0, uN =zN, and z¯ is a subsequence of u¯.

Proof. We first remark that if x, y ∈ A˜, then the sequence (x, y,y, x, y, x,x, y) is a simple 1 -sequence, called a trick sequence. It has three steps, and it joins x, y to itself.

The theorem is proved by induction on N. Assume first that N = 1 . Then

¯

z is a pair (z, z) with (z, z) ≤λ. Write z =x, y= (x, r) and z =x, y= (x, r) . I f x = x, we can use two trick sequences to embed ¯z into a simple λ- sequence with 7 steps. We may thus assume that |x−x|=t >0 . By symmetry, we may assume that r ≥r. We consider two cases.

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Case 1. r ≤eλt. Consider the sequence ¯v = (z, v1, v2, z) , where v1 =x, x, v2 = x, x. I f r t, then (z, v1) = log(r/t) λ. I f r t, then 2.4 and 2.7 imply that

(z, v1)< (z, v1) +(v1, v2) =(z, v2) =

z, L(x)

≤(z, z) ≤λ.

A similar argument shows that (v2, z)≤λ. Hence ¯v is a λ-sequence. Extending

¯

v by two trick sequences we obtain the desired simple λ-sequence ¯u from z to z. Case 2. r eλt. We show that the sequence ¯v = (z, v1, v2, v3, z) with v1 =y, x, v2 =y, x, v3 =x, y is a λ-sequence.

Since

(4.4) r−t ≤ |y−x| ≤r+t,

we obtain

(v1, v2) =log r

|y−x| log r

r−t log e

e−1 <1.

To prove that (v3, z)≤λ we consider two subcases.

Subcase 2a. r ≤r−t. By (4.4) and by 2.4 we obtain (v3, z) = log|y−x|

r log r

r + log 1 + t r

log r r + t

r =(z, z)≤λ.

Subcase 2b. r ≥r−t. Since r≥eλt, it follows from (4.4) that the numbers r and |y−x| are between r(1−eλ) and r(1 +eλ) . Hence

(v3, z)log 1 +eλ

1−eλ loge+ 1 e−1 <1.

We have proved that ¯v is a λ-sequence. Extending ¯v by a trick sequence from z to z we obtain a simple λ-sequence from z to z.

Next assume that p is a positive integer and that the theorem is true whenever N p. Suppose that N = p+ 1 . Applying the induction hypothesis to the λ- sequences (z0, . . . , zN1) and (zN1, zN) we embed them into simple λ-sequences.

Linking these sequences with the trivial vertical step (zN1, zN1) we obtain the desired simple λ-sequence ¯u.

4.5. Remark. The construction of the proof of 4.3 is not the most economical.

It gives the bound N 10N.

4.6. Relative connectedness. We say that a finite sequence ¯x = (x0, . . . , xN) in a metric space A is proper if N 1 and if xj1 = xj for all 1 j ≤N. I n particular, a pair (x, y) is proper if x =y. The sequence ¯x is called M-relative, M 1 , if ¯x is proper and if

|xj1 −xj|/M ≤ |xj −xj+1| ≤M|xj1−xj|

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or equivalently,

log |xj−xj+1|

|xj1−xj|≤logM

for all 1 j N 1 . In the trivial case N = 1 , ¯x is a pair (x0, x1) , and the condition for M-relativity is vacuously true for all M 1 .

We say that the sequence ¯x = (x0, . . . , xN) joins the pairs (x0, x1) and (xN1, xN) , and that A is M-relatively connected if each pair of proper pairs (x, y) and (x, y) in A can be joined by an M-relative sequence in A. The space A is relatively connected if it is M-relatively connected for some M 1 .

A connected space is M-relatively connected for all M >1 , but so are many other spaces as well, for example, the Cantor middle-third set.

4.7. Associated sequences. We shall show that the eλ-relative sequences of A are in one-to-one correspondence with the simple λ-sequences of the upper set ˜A. Suppose that ¯x = (x0, . . . , xN) is a proper sequence in A. Define a sequence

¯

z = (z0, . . . , z2N−1) in ˜A by

z2j =xj, xj+1, z2j+1 =xj+1, xj

for 0 j ≤N 1 . Thus ¯z is the sequence

(x0, x1,x1, x0,x1, x2,x2, x1, . . . ,xN−1, xN,xN, xN−1).

Clearly ¯z is a simple sequence. Moreover, ¯z is a λ-sequence if and only if ¯x is eλ-relative; remember that we always assume that λ≥1 . We write ¯z = As ¯x and say that ¯z is the simple sequence associated with ¯x.

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Q1

Q2

Q3

x0

x1

x2 x3

z0

z1

z2

z3

z4 z5

Rn

Figure 2. ¯z= As ¯x

Conversely, if ¯z = (z0, . . . , z2N1) is a simple sequence in ˜A, we associate with ¯z the proper sequence ¯x = As1z¯= (x0, . . . , xN) , where x0 =π(z0) , xN = π(z2N1) , and xj = π(z2j1) = π(z2j) for 1 j N 1 . Then As As−1z¯= ¯z and As1As ¯x= ¯x for all ¯z and ¯x. We have proved:

4.8. Theorem. The function x¯ As ¯x gives a bijective correspondence between the proper sequences of A and the simple sequences of A˜. Moreover, x¯ is eλ-relative if and only if As ¯x is a λ-sequence.

4.9. Theorem. For λ 1, a metric space A is eλ-relatively connected if and only if A˜ is λ-connected.

Proof. This follows from 4.3 and 4.8.

4.10. Remark. I n the case A⊂Rn, we can find As ¯x by erecting on each line segment [xj1, xj] a square Qj, orthogonal to Rn (identified with Rn × {0} ⊂ Rn+1) . The sequence ¯z = As ¯x consists of vertices of these squares as shown in Figure 2.

We next give an alternative characterization for relative connectedness.

4.11. Theorem. For a metric space A, the following conditions are quanti- tatively equivalent:

(1) A is M-relatively connected.

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(2) There is c≥1 such that if x ∈A andB(x, r) =A, then eitherB(x, r) = {x} orB(x, r)\B(x, r/c)= ∅.

More precisely, (1) implies (2) with c= 2M + 1, and (2) implies (1)with all M > c.

Proof. Assume that A is M-relatively connected and that (2) is not true for c= 2M + 1 3 . Then there are x and r such that {x} =B(x, r) =A and such thatB(x, r)\B(x, r/c) =∅. Pick points y and y in A such that

0<|x−y|< r/ c, |x−y|> r.

Since A is M-relatively connected, there is an M-relative sequence (x0, . . . , xN) joining (x, y) to (x, y) . Let k be the smallest number with |xk−x| ≥ r. Then k 2 , and

M |xk−xk1|

|xk1−xk2| > r−r/c

2r/c = c−1 2 , which gives c <2M + 1 , a contradiction.

Conversely, assume that (2) is true, that M > c, and that (x, y) and (x, y) are proper pairs in A. We show that they can be joined by an M-relative sequence.

We first assume that x= x. By symmetry, we may assume that |x−y| ≤ |x− y|. Applying condition (2) inductively, we find a sequence of points y =y0, . . . , yN

such that

|x−yj|/M ≤ |x−yj+1| ≤c|x−yj|/M

for 0 ≤j ≤N−1 and |x−yN|/M ≤ |x−y| ≤ |x−yN|. Now (x, y0, x, y1, . . . , x, yN, x, y) is an M-relative sequence from (x, y) to (x, y) .

If x = x, we can join the pairs (x, y) and (x, y) to (x, x) and (x, x) , respectively, by c-relative sequences. These can be joined in a natural way to a c-relative sequence from (x, y) to (x, y) .

4.12. Uniformly perfect spaces. A metric space A is c-uniformly perfect if B(x, r) =A implies thatB(x, r)\B(x, r/c)=∅. The concept was introduced by C. Pommerenke [Po] for closed unbounded sets in R2, and it has turned out to be useful in various questions in analysis.

A uniformly perfect set containing more than one point has no isolated points.

For example, a finite metric space containing more than one point is not uniformly perfect although it is relatively connected. The following corollary of 4.11 gives relations between uniform perfectness and relative connectedness.

4.13. Theorem. A c-uniformly perfect space is M-relatively connected for all M > c. If a space A is M-relatively connected and has no isolated points, then A is c-uniformly perfect with c= 2M + 1.

We apply the results of this section to show that if points z < z can be joined by a λ-sequence in ˜A, they can be joined by a vertical (λ+ 1) -sequence. Recall the notation L(x) ={x} ×(0,) from 2.6.

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4.14. Lemma. Suppose that z, z ∈γ Γλ(A) and that π(z) = π(z) =x. Then z and z can be joined by a (λ+ 1)-sequence in γ∩L(x).

Proof. Write z = x, y, z = x, y, and assume, for example, that z < z. By 4.3 and 4.8, there is an eλ-relative sequence (x0, . . . , xN) in A joining the pairs (x, y) and (x, y) . For 1 i N we set pi = max{|xk−x| : 1 k i}. Then

|xi−xi−1| ≤ |xi −x|+|x−xi−1| ≤2pi, and hence

|xi+1−x| ≤ |xi+1−xi|+|xi−x| ≤2pieλ+pi, which yields pi+1 (2eλ+ 1)pi. Consequently,

log pi+1

pi log(2eλ+ 1) =λ+ log(2 +eλ)< λ+ 1.

For each i [1, N] choose j(i) i with |xj(i) x| = pi, and set zi = x, xj(i). Then zi A˜, z =z1 z2 ≤ · · · ≤ zN, and z zN. Choose k with zk ≤z ≤zk+1. Since

(zi, zi+1) = logpi+1

pi < λ+ 1,

(z1, . . . , zk, z) is a (λ+ 1) -sequence joining z and z in ˜A∩L(x) . Since z, z ∈γ, all zi lie in γ by 3.4(6).

We apply 4.14 to prove the following more general result.

4.15. Theorem. If z = (x, r) and z = (x, r) are in a λ-component γ, then z and z can be joined by a (λ+ 1)-sequence in γ∩

L(x)∪L(x) .

Proof. By 4.14 we may assume that x=x. Choose β Γ containing x, x and x, x. I f β = γ, then β < γ by 3.4(9). Choose y A with z = x, y. Writing z1 =x, y we have (z, z1)≤λ by 3.4(1). Then z1 γ, and z can be joined to z1 by a (λ+ 1) -sequence in γ∩L(x) by 4.14.

5. Order properties of Γλ(A)

5.1. Summary. We continue the study of the set Γ = Γλ(A) , especially from the order-theoretic point of view. To this end, we introduce the order-theoretic concept of a family tree and show that Γ has this property.

5.2. Family trees. We consider an arbitrary partially ordered set (P,≤) . We recall that P is (upwards) directed if for each pair x, y ∈P there is z P with x z y. I f P is directed, then a maximal element in P is also the greatest element of P, written as maxP if it exists.

We say that P is a family tree if

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(1) P is directed,

(2) for each pair a, b P, the set [a, b] = {x ∈P :a x b} is finite and linearly ordered.

Suppose that P is a family tree and that a P, a = maxP. We show that the set P+(A) = {x P : x > a} has the least element, written as p(a) . Choosing b∈ P with b > a we can write [a, b] = {x0, . . . , xN}, where a = x0 <

x1 < · · · < xN = b. I f y > a, there is z P with y z x1. Since [a, z] is linearly ordered, we have either x1 ≤y or x1 > y. The second case is impossible, since [a, x1] ={a, x1}. Hence x1 is the least element of P+(a) .

If b = p(a) , we say that b is the parent of a and a is a child of b. An element a = maxP has precisely one parent, but each element of P may have several children. Writing p2(a) =p

p(a)

etc., we have (5.3) P+(a) ={p(a), p2(a), p3(a), . . .}.

The sequence may be infinite or finite. The latter case occurs if and only if maxP exists; then the sequence ends with maxP. We see that P+(a) is always linearly ordered.

A family tree is obviously a semilattice, that is, each pair a, b P has the least upper bound a∨b.

We remark that the Hasse diagram of a family tree is a tree in the graph- theoretic sense: it is connected and contains no cycles.

We now return to the family Γ = Γλ(A) of all λ-components of the upper set ˜A of a metric space A. We recall from 3.2 and from 3.4 the following three equivalent characterizations for the ordering γ ≤γ of elements of Γ :

(1) πγ ⊂πγ.

(2) z ≤z for some z ∈γ and z ∈γ.

(3) For each z ∈γ there is z ∈γ with z ≤z. 5.4. Theorem. Γλ(A) is a family tree.

Proof. To show that Γ is directed, assume that γ and γ are incomparable elements of Γ . Choose x, y ∈γ and x, y γ. Then x, y ∈πγ. By 3.4(10), we have |x −x| ≥ eλ|x −y| > |x −y|, and hence x, y ≤ x, x. Similarly x, y ≤ x, x. Since there is γ Γ containing x, x and x, x, Γ is directed.

Next assume that γ, γ Γ with γ < γ. We can choose points (x, r) γ and (x, r) γ with r < r. I f β Γ with γ < β < γ, then β contains a point (x, s) with r < s < r. Hence [γ, γ] is linearly ordered. Moreover, [γ, γ] is finite, since |log(s/s)| > λ whenever (x, s) and (x, s) belong to different λ-components of ˜A.

5.5. Remark. It follows from 5.4 that each γ Γ with γ = max Γ has a parent p(γ) . To find p(γ) , choose an arbitrary point (x, r) γ. Then (x, r)

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p(γ) whenever (x, r) A˜ and t r < teλ, where t = inf{s : r < s, (x, s) A˜\γ}.

In the next result we give an alternative way to find p(γ) . Observe that γ = max Γ if and only if πγ =A.

5.6. Lemma. Let γ Γ, and let x πγ, y A \πγ be such that

|x−y| ≤ed(πγ, A\πγ). Then x, y ∈p(γ).

Proof. Choose y πγ and γ Γ such that x, y ∈ γ and x, y γ. By 3.4(10) we have |x−y| > eλ|x−y|, and hence x, y <x, y. This implies γ < γ, and thus p(γ)≤γ. I f p(γ)= γ, then p(γ) contains a point x, y such that

|x−y|< e−λ|x−y| ≤e1−λd(πγ, A\πγ)≤d(πγ, A\πγ).

By 3.4(1) we have y ∈A\πγ. Since x∈πγ, this gives a contradiction.

5.6. Lemma. Let γ Γ, and let x, y∈πγ, z ∈A\πγ. Then

|x−y| ≤e(γ,p(γ))|x−z|.

Proof. Let β and γ be the members of Γ containing x, y and x, z, respectively. By 3.4(8), we have β γ. Since x πγ ∩πγ and since z /∈ πγ, we have γ < γ by 3.4(4). Hence p(γ) γ. By 3.4(3), there is w p(γ) with x, y < w≤ x, z. Consequently,

γ, p(γ)

≤(x, y, w)≤log |x−z|

|x−y|, and the lemma follows.

The following order-theoretic result is needed in Section 6.

5.8. Lemma. Let (P,) be a family tree, and let g: P [0,) be an unbounded function. Then there is a sequence (xj) in P such that g(xj) → ∞ and such that one of the following three conditions is satisfied:

(1) x1 > x2 >· · ·; (2) x1 < x2 <· · ·;

(3) xi and xj are incomparable for all i=j. Proof. For x∈P we set

h(x) = sup{g(y) :y < x} ∈[0,]

with the agreement that h(x) = 0 for all minimal elements x of P. We consider three cases.

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Case 1. h(x)<∞ for all x ∈P. Now maxP does not exist. Fix an arbitrary y1 ∈P and set yn+1 =pn(y1) for n∈N. Then yn < yn+1 and h(yn)≤h(yn+1) for all n. We first show that h(yn) → ∞. Let M >0 , and choose x P with g(x)> M. Setting y =p(x∨y1) we have y∈P+(y1) , and hence y =yn for some n by (5.3). Thus h(yn)≥g(x)> M.

I f sup{g(yn) : n N} = , there is a subsequence (xn) of (yn) such that g(xn) → ∞ and (2) is true. Suppose that sup{g(yn) : n N} = c < . Since h(yn) → ∞, there is a sequence n0 < n1 < n2 < · · · in N such that h(yn0) > c and h(yni+1) > h(yni) for all i 0 . Choose elements x1, x2, . . . in P such that xi+1 < yni+1 and g(xi+1)> h(yni) for all i 0 . Then g(xi)→ ∞. It suffices to show that xj and xi+1 are incomparable whenever 1≤j ≤i.

If xi+1 ≤xj, then xi+1 < ynj, and we obtain the contradiction h(yni)< g(xi+1)≤h(ynj)≤h(yni).

Assume that xj < xi+1. Now xi+1 and ynj are in P+(xj) . Since this is linearly ordered by (5.3), either xi+1 < ynj or ynj xi+1. As above, the first case is impossible. In the second case, xi+1 =yk for some k, and we get the contradiction

c < h(yn0)≤h(yni)< g(xi+1) =g(yk)≤c.

Case 2. There is y P such that h(y) = and such that h(z) < for every child z of y. Let C = p−1{y} be the set of all children of y. I f g is unbounded in C, we choose a sequence (xn) in C with g(xn) → ∞; then (xn) satisfies (3). If g is bounded in C, then h is unbounded in C, and there is a sequence z0, z1, . . . in C such that h(zi+1)> h(zi) and h(zi) → ∞. For i≥0 we choose elements xi+1 < zi+1 with g(xi+1) > h(zi) . Now g(xi) → ∞. For i = j we have xi∨xj =y, and hence xi and xj are incomparable.

Case 3. Cases 1 and 2 do not occur. Then there is a sequence y1, y2, . . . in P such that h(yn) = and yn =p(yn+1) for all n∈N. I f sup{g(yn) :n∈N}=, then (yn) has a subsequence (xn) such that g(xn)→ ∞ and (1) holds. Suppose that

sup{g(yn) :n∈N}=c <∞.

We define a sequence n1 < n2 < · · · in N and elements xi < yni inductively as follows. Set n1 = 1 and choose x1 < y1 with g(x1) > c. Assume that ni and xi < yni have been defined. Since [xi, yni] is finite, there is ni+1 > ni such that xi is not smaller than yni+1. Choose xi+1 < yni+1 such that g(xi+1)> g(xi) + 1 . Now g(xi) → ∞, and it suffices to show that xj and xi+1 are incomparable whenever 1≤j ≤i.

If xj ≤xi+1, then xj < yni+1 ≤ynj+1, a contradiction. If xj > xi+1, then xj and yni+1 are in the linearly ordered set P+(xi+1) , and hence either xj < yni+1

or xj ≥yni+1. In the first case we again get the contradiction xj < ynj+1. I n the second case, xj [yni+1, y1] , and hence xj = yk for some k. This is impossible, since g(yk)≤c < g(xj) .

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